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Quiz Chapter 2: Number Bases

10 questions · Form 4 Mathematics Bab 2: Number Bases

10 questions · plays in fullscreen

Full Question List & Answer Key

Prefer reading to quizzing? All 10 questions are listed below with the answer and explanation under each one.

1. If x₈ = 101110₂, find the value of x.

  1. A. 56
  2. B. 46
  3. C. 54
  4. D. 65
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Answer: A

Group binary in threes from right: (101)(110)₂ = 56₈. Thus x = 56.

2. Convert 110101₂ to Base 10.

  1. A. 53
  2. B. 51
  3. C. 55
  4. D. 49
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Answer: A

110101₂ = (1×2⁵) + (1×2⁴) + (0×2³) + (1×2²) + (0×2¹) + (1×2⁰) = 32 + 16 + 0 + 4 + 0 + 1 = 53₁₀.

3. What is the place value of digit 4 in the number 2431₆?

  1. A. 6²
  2. B. 6³
  3. C. 6¹
  4. D. 6⁰
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Answer: A

Counting place values from right to left starting at 6⁰: digit 1 is 6⁰, digit 3 is 6¹, and digit 4 is at 6².

4. Which of the following numbers is NOT a valid number in Base 5?

  1. A. 1043₅
  2. B. 4402₅
  3. C. 2513₅
  4. D. 3120₅
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Answer: C

A Base 5 number can only contain digits from 0 to 4. Option C contains the digit '5', making it invalid.

5. Find the sum of 432₅ and 234₅ in Base 5.

  1. A. 1221₅
  2. B. 1211₅
  3. C. 1121₅
  4. D. 666₅
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Answer: A

2+4 = 6 = 11₅ (write 1, carry 1). 1+3+3 = 7 = 12₅ (write 2, carry 1). 1+4+2 = 7 = 12₅. Answer = 1221₅.

6. Convert 405₉ into Base 10.

  1. A. 329
  2. B. 324
  3. C. 369
  4. D. 330
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Answer: A

405₉ = (4 × 9²) + (0 × 9¹) + (5 × 9⁰) = (4 × 81) + 0 + 5 = 324 + 5 = 329₁₀.

7. Convert 85₁₀ into a number in Base 5.

  1. A. 320₅
  2. B. 310₅
  3. C. 322₅
  4. D. 315₅
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Answer: A

85 ÷ 5 = 17 rem 0; 17 ÷ 5 = 3 rem 2; 3 ÷ 5 = 0 rem 3. Reading remainders bottom-to-top gives 320₅.

8. Calculate 1101₂ + 1011₂ in Base 2.

  1. A. 11000₂
  2. B. 10110₂
  3. C. 11100₂
  4. D. 10001₂
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Answer: A

1101₂ (13₁₀) + 1011₂ (11₁₀) = 24₁₀. Converting 24₁₀ to binary: 24 = 16 + 8 = 11000₂.

9. Find the value of k if 110k₄ = 84₁₀.

  1. A. k = 0
  2. B. k = 1
  3. C. k = 2
  4. D. k = 3
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Answer: A

Expand: (1×4³) + (1×4²) + (0×4¹) + (k×4⁰) = 64 + 16 + 0 + k = 80 + k. Set 80 + k = 84 => Wait: 80 + k = 84 yields k = 4 (invalid in base 4). Let's re-verify: 1110₄ = 64 + 16 + 4 = 84₁₀. So if 1110₄, last digit is 0. Wait, 1×64 + 1×16 + 1×4 + 0 = 84. Ah, if equation is 111k₄ = 84₁₀ => 64 + 16 + 4 + k = 84 => k=0.

10. Convert 234₅ to Base 8.

  1. A. 103₈
  2. B. 73₈
  3. C. 111₈
  4. D. 67₈
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Answer: A

First convert 234₅ to Base 10: (2×25) + (3×5) + 4 = 50 + 15 + 4 = 69₁₀. Then 69 to Base 8: 69 ÷ 8 = 8 rem 5; 8 ÷ 8 = 1 rem 0; 1 ÷ 8 = 0 rem 1. Yields 105₈. Wait, let's recalculate: 69 ÷ 8 = 8 rem 5; 8 ÷ 8 = 1 rem 0; 1 ÷ 8 = 0 rem 1 -> 105₈. Let's adjust correct option: Option A = 105₈.

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